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Giant Fullerene with 540 atoms, 12 pentagons, 260 hexagons

A general rule for the number of atoms (n) in such a large fullerene is to count the number of bonds (a) that lie between the corners of the pentagons to the next pentagon (in this case a=3) square this number and times by 60 (i.e. n=60xaxa, 3x3x60=540 atoms). The number of bonds (b) is 1.5 times the number of atoms (i.e. b=1.5xn, 1.5 x 540 = 810 bonds). The number of hexagons (h) is give by h=(n-20)/2=(540-20)/2=260 hexagons.
(picture, Ken McKay, The Sussex Fullerene Centre)


Dr Jonathan Hare, Room 3R253, Chichester Bldg. CPES, The University of Sussex
Brighton, East Sussex. BN1 9QJ. 01273 606755 x3171

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